[Loael]

Which of the following mixtures creates a buffer solution?

A) 10.0 ml of 0.10M NaOH, 10.0ml of 0.10M HF
B) 20.0 ml of 0.10M NaOH, 15.0ml of 0.10M HF
C) 15.0 ml of 0.10M NaOH, 20.0ml of 0.10M HF
D) 10.0 ml of 0.10M NaOH, 5.0ml of 0.20M HF

Well I new that A and D were not the answers because they are essentially the same thing. The answer is C but I'm not exactly sure why.



Consider the system at equilibrium:
2SO2(g) + O2(g)  <--> 2SO3(g)

for which (delta)H < 0. Which change(s) will increase the yield of SO3(g)?

I. Increasing the temperature
II. Increasing the volume of the container

A) I only
B) II only
C) Both I and II
D) Neither I nor II



For this one, I new that the second statement would not be true because that would decrease the pressure, which favors the side with more moles of gas, in this case the reverse reaction. So I cancelled out choices B and C. But I'm not very clear on how temperature affects an endothermic reaction. I put A but the answer was D. If this reaction was exothermic, would the answer be A?







How much heat is required to convert 5.0g of ice at -10.0 C to liquid water at 15.0 C? (assume heat capacities are independent of temperature.)

Enthalpy of fusion: 6.00 kJ/mol
Specific heat capacity of ice: 37.8 J/mol
Specific heat capacity of water: 76.0 J/mol


A) 4.2 x 10^2 J
B) 2.1 x 10^3 J
C) 9.3  10^3 J
D) 3.8 x 10^4 J


I'm in AP Chem right now and my teacher never really brushed over heat capacities at all. I put D, but the answer was B. Can someone help me out on this?

[Borek]

1. What is definition of a buffer solution? What must it contain?

2. Think in terms of LeChatelier's principle.

3. You have to heat up the ice, melt it, and heat up the water. Can you calculate amount of heat for each step?

[Loael]

1.) Well it's a weak acid/base with its conjugate base/acid. Is it C because not all of the NaOH reacts with the HF?

2.) I feel retarded lol. I do not know why I said it was endothermic. But yes now I understand, increasing the temperature for this would want the reaction to form more of the reactants so that it can use the outside temperature so that the reaction can proceed, right?

3.) For this I did:

(5g ice)(10 C)(37.8 J/mol) = 1890 J/mol
(6000J/mol)(5g) = 30000J/mol
(5g water)(15 C)(76 J/mol) = 5700 J/mol

add them all together is 37590 J/mol which looks like D. I'm still not sure how it's B.

[Borek]

(5g ice)(10 C)(37.8 J/mol) = 1890 J/mol

Units, my son.

[Loael]

(5g h2o / 18.02 g) = .2775 mol h2o


(.2775 mol H2O)(37.8 J/mol)(10 C) = 104.895 J*C

(.2775 mol H2O)(76 J/mol)(15 C) = 316.35 J*C

(.2775 mol H2O)(6 kJ/mol) = 1665 J*C

Add them up is 2086 J

I'm still doing the same thing, but I did go look back in my Chem I binder and did it with these values:

(4.184 J/G*C)*(5g water)*(15 C) = 313.8 J
(2.06 J/G*C) * (5g ice) * (10 C) = 103 J
334 J/g * 5grams water = 1670 J

add these together I get 2080J. The same value



Thank you though for the help. And did I explain the other two questions correctly?